Question

Suppose that in a random selection of 100 colored candies, 21% of them are blue. The candy company claims that the percentage of blue candies is equal to 29%. Use the 0.01 significance level to test that claim.

Answer 1

The pvalue of the test is 0.0784 > 0.01, which means that at this significance level, we do not reject the null hypothesis that the percentage of blue candies is equal to 29%.

Explanation:

The candy company claims that the percentage of blue candies is equal to 29%.

This means that the null hypothesis is:

`H_(0): p = 0.29`

We want to test the hypothesis that this is true, so the alternate hypothesis is:

`H_(a): p \\eq 0.29`

The test statistic is:

`z = (X - \mu)/((\sigma)/(√(n)))`

In which X is the sample mean,
`\mu` is the value tested at the null hypothesis,
`\sigma` is the standard deviation and n is the size of the sample.

0.29 is tested at the null hypothesis:

This means that
`\mu = 0.29, \sigma = √(0.29*0.71)`

Suppose that in a random selection of 100 colored candies, 21% of them are blue.

This means that
`n = 100, X = 0.21`

Value of the test statistic:

`z = (X - \mu)/((\sigma)/(√(n)))`

`z = (0.21 - 0.29)/((√(0.29*0.71))/(√(100)))`

`z = -1.76`

Pvalue of the test:

The pvalue of the test is the probability of the sample proportion differing at least 0.21 - 0.29 = 0.08 from the population proportion, which is 2 multiplied by the pvalue of Z = -1.76.

Looking at the z-table, z = -1.76 has a pvalue of 0.0392

2*0.0392 = 0.0784

The pvalue of the test is 0.0784 > 0.01, which means that at this significance level, we do not reject the null hypothesis that the percentage of blue candies is equal to 29%.